Differentiation Rules
Differentiation rules are formulae that allow us to find the derivatives of functions quickly.
Contents
- Basic Properties of Derivatives
- Derivatives of Polynomials (Power Rule)
- Derivatives of Trigonometric Functions
- Derivatives of Exponential Functions
- Derivatives of Logarithmic Functions
- Chain Rule
- Product Rule
- Quotient Rule
- Inverse Functions
- Differentiation Rules Problem Solving - Basic
- Differentiation Rules Problem Solving - Intermediate
- Differentiation Rules - Problem Solving - Advanced
Basic Properties of Derivatives
Taking derivatives of functions follows several basic rules:
- multiplication by a constant: \( \big(c\cdot f(x)\big)' = c \cdot f'(x) \)
- addition and subtraction: \( \big( f(x)\pm g(x)\big)' = f'(x) \pm g'(x).\)
For multiplication and composition of functions, see product rule and chain rule.
Derivatives of Polynomials (Power Rule)
Main article: Power Rule
When taking the derivatives of polynomials, we can use the power rule:
Power Rule
\[ \frac{d}{dx} x^n = n\cdot x^{n-1} \]
Derivatives of Trigonometric Functions
Main Article: Differentiation of Trigonometric Functions
We can see the basic trigonometric derivatives in the table below:
| \( f(x) \quad \rightarrow \) | \( f'(x) \) |
| \( \sin x \) | \( \cos x \) |
| \( \cos x \) | \( -\sin x \) |
| \( \tan x \) | \( \sec^2 x \) |
| \( \sec x \) | \( \sec x \tan x \) |
| \( \csc x \) | \( -\csc x \cot x \) |
| \( \cot x \) | \( - \csc^2 x \) |
Derivatives of Exponential Functions
Main Article: Differentiation of Exponential Functions
The main formula you have to remember here is the derivative of a logarithm:
\[\dfrac{d}{dx} \log_a x = \dfrac{1}{x \cdot \ln a}.\]
What is the derivative of the following exponential function:
\[ f(x) = 2 \cdot 3^x ?\]
We have
\[ \begin{align} f'(x) &= 2(3^x)' \\ &= 2\cdot 3^x \ln 3\\ &= 2(\ln 3)\cdot 3^x. \ _\square \end{align} \]
What is the derivative of the following exponential function:
\[ f(x) = 3e^x ?\]
We have
\[ \begin{align} f'(x) &= 3(e^x)' \\ &= 3e^x. \ _\square \end{align} \]
Derivatives of Logarithmic Functions
Main Article: Differentiation of Logarithmic Functions
What is the derivative of the following logarithmic function:
\[ f(x) = \ln x + x ?\]
We have
\[ \begin{align} f'(x) &= (\ln x)' +x' \\ &= \frac{1}{x} + 1. \ _\square \end{align} \]
What is the derivative of the following logarithmic function:
\[ f(x) =3\log_{2} x -x^2 +2 ?\]
We have
\[ \begin{align} f'(x) &= (3\log_{2}x)'-\big(x^2\big)' +0 \\ &= 3\cdot \frac{1}{x} \cdot \frac{1}{\ln 2} - 2x \\ &= \frac{3}{x\ln 2} - 2x. \ _\square \end{align} \]
Chain Rule
Main Article: Chain Rule
General form : \(\frac{d}{dx} f\big(g(x)\big) = f'\big(g(x)\big) \cdot g'(x)\)
What is the derivative of the following function:
\[ y=f(2x^2+x-3)?\]
We have
\[ \begin{align} y' &= f'(2x^2+x-3)\big(2x^2+x-3\big)' \\ &= (4x+1)f'(2x^2+x-3). \ _\square \end{align} \]
What is the derivative of the following function:
\[ f(x) = (x^2+x+1)^5 ?\]
We have
\[ \begin{align} f'(x) &= 5\big(x^2+x+1\big)^4\big(x^2+x+1\big)' \\ &= 5\big(x^2+x+1\big)^4(2x+1). \ _\square \end{align} \]
What is the derivative of the following function:
\[ y = \sqrt{2x+1} ?\]
Since \( y = \sqrt{2x+1} = (2x+1)^{\frac{1}{2}} ,\) we have
\[ \begin{align} y' &= \frac{1}{2}(2x+1)^{\frac{1}{2}-1} \cdot (2x+1)' \\ &= (2x+1)^{-\frac{1}{2}} \\ &= \frac{1}{\sqrt{2x+1}}. \ _\square \end{align} \]
Product Rule
Main Article: Product Rule
What is the derivative of the following function:
\[ y = (x^2+3)(x^3+2x^2+5) ?\]
We have
\[ \begin{align} y' &= (x^2+3)'(x^3+2x^2+5) + (x^2+3)(x^3+2x^2+5)' \\ &= 2x(x^3+2x^2+5) + (x^2+3)(3x^2+4x) \\ &= 5x^4 + 8x^3 + 9x^2 +22x. \ _\square \end{align} \]
What is the derivative of the following function:
\[ y=(x^2+1)(2x+1)(3x+1) \]
when \(x = 1?\)
We have
\[ \begin{align} y' &= (x^2+1)'(2x+1)(3x+1) + (x^2+1)(2x+1)'(3x+1) + (x^2+1)(2x+1)(3x+1)' \\ &= 2x(2x+1)(3x+1) + (x^2+1)\cdot 2 \cdot (3x+1) + (x^2+1)(2x+1) \cdot 3 \\ &= 24x^3 + 15x^2 + 14x +5. \end{align} \]
Thus, the derivative of \(y\) at \(x=1\) is
\[y' \big\vert_{x=1} =24 \cdot 1^3 + 15 \cdot 1^2 + 14 \cdot 1 + 5 = 58. \ _\square\]
Quotient Rule
Main Article: Quotient Rule
What is the derivative of the following function:
\[ y = \frac{1-x}{x^2+2} ?\]
We have
\[ \begin{align} y' &= \frac{(1-x)'(x^2+2)-(1-x)(x^2+2)'}{(x^2+2)^2} \\ &= \frac{(-1) \cdot (x^2+2)-(1-x) \cdot 2x}{(x^2+2)^2} \\ &= \frac{x^2-2x-2}{(x^2+2)^2}. \ _\square \end{align} \]
What is the derivative of the following function:
\[ y = \frac{1}{x^2\sqrt{x^3}} ?\]
Since \(y= \frac{1}{x^2\sqrt{x^3}} = x^{-\frac{7}{2}}, \) we have \[ \begin{align} y' &= -\frac{7}{2} x^{-\frac{9}{2}} \\ &= -\frac{7}{2x^4\sqrt{x}}. \ _\square \end{align} \]
Inverse Functions
Main Article: Differentiation of Inverse Functions
If \(f(x) = x^3+ 7x+2,\) find \((f^{-1})'(10).\)
If \(f\) is a one-to-one differentiable function with inverse function \(f^{-1}\) and \(f'\big(f^{-1}(a)\big) \neq 0,\) then the inverse function is differentiable at \(a\) and
\[(f^{-1})'(a)=\frac{1}{f'\big(f^{-1}(a)\big)}.\]
Notice that \(f\) of this problem is one-to-one because
\[f'(x)=3x^2+7 > 0,\]
so \(f\) is increasing. To use the above theorem, we need to know \(f^{-1}(10)\) and we can find it by equating \(f(x)=10:\)
\[ \begin{align} 10 &= x^3 + 7x + 2 \\ \Rightarrow x &= 1, f^{-1}(10) = 1. \end{align} \]
Therefore,
\[(f^{-1})'(10) = \frac{1}{f'\big(f^{-1}(10)\big)} = \frac{1}{f'(1)} = \frac{1}{3\cdot 1^2 + 7} = \frac{1}{10}. \ _\square\]
Differentiation Rules Problem Solving - Basic
If \( f(x+y) = f(x) + f(y) + xy \) and \(f'(0)=2,\) what is \(f'(2)?\)
Substituting \(x=0\) and \(y=0\) into \( f(x+y) = f(x) + f(y) + xy ,\) we get the value of \(f(0)\) as follows:
\[ f(0) = f(0) + f(0) + 0 \implies f(0) = 0 .\]
Thus,
\[ \begin{align} f'(2) &= \lim_{h\rightarrow 0} \frac{f(2+h)-f(2)}{h} \\ &= \lim_{h \rightarrow 0} \frac{f(2) + f(h) + 2h - f(2)}{h} \\ &= \lim_{h \rightarrow 0} \left ( \frac{f(h)}{h}+2\right ) \\ &= \lim_{h \rightarrow 0} \left ( \frac{f(0+h)-f(0)}{h}+2\right ) \\ &= f'(0) + 2 \\ &= 2+2 \\ &= 4. \ _\square \end{align} \]
Differentiation Rules Problem Solving - Intermediate
If \(f(x) = \left(x + \sqrt{1+x^2}\right)^{10},\) what is \( f'(1)\cdot f'(-1) ?\)
We have
\[ \begin{align} f'(x) &= 10\left(x+\sqrt{1+x^2}\right)^9 \times \left(x+\sqrt{1+x^2}\right)' \\ &= 10\left(x+\sqrt{1+x^2}\right)^9\left(1 + \frac{2x}{2\sqrt{1+x^2}} \right ) \\ &= \frac{10}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^{10}. \end{align} \]
Therefore,
\[ \begin{align} f'(1) \cdot f'(-1) &= \frac{10}{\sqrt{2}} \left(1+\sqrt{2}\right)^{10} \times \frac{10}{\sqrt{2}}\left(-1 + \sqrt{2}\right)^{10} \\ &= 50\left(\big(\sqrt{2} +1\big)\big(\sqrt{2} -1\big)\right)^{10} \\ &= 50(2-1)^{10} \\ &= 50. \ _\square \end{align} \]
Differentiation Rules - Problem Solving - Advanced
If \( F(t) = f\big(g(t)\big), f'(t) = \frac{1}{t^2+1}, g'(t) = \frac{10}{t^4 + 1},\) and \(g(0) = 3,\) what is \( F'(0) ?\)
Since \(F'(t) = f'\big(g(t)\big)g'(t),\)
\[ \begin{align} G'(0) &= f'\big(g(0)\big)g'(0) \\\\ &= f'(3)\cdot \frac{10}{0^4+1} \\\\ &= \frac{1}{3^2 +1}\cdot 10 \\\\ &= 1. \ _\square \end{align} \]
The function \(F(x)\) is defined by the following identity:
\[F\left(\big(F(x)+x\big)^k\right) = \big(F(x)+x\big)^2-x.\]
The value of \(F(1)\) is such that a finite number of possible numerical values of \(F'(1)\) can be determined solely from the above information. The maximum value of \(k\) such that \(F'(1)\) is an integer can be expressed as \(\frac{a}{b}\), where \(a\) and \(b\) are coprime integers.
What is the value of \(a+b?\)
What is the derivatives of following function:
\[ y= \sqrt[3]{(x+1)(x^2+1)}\] when \(x=0 ?\)
We have
\[ \begin{align} y &= \sqrt[3]{(x+1)(x^2+1)} \\ \Rightarrow y^3 &= (x+1)(x^2+1). \end{align} \]
Differentiating both sides with respect to \(x,\) we have
\[ \begin{align} 3y^2 \frac{dy}{dx} &= (x^2+1) +(x+1) \cdot 2x \\ &= 3x^2 +2x + 1 \\\\ \Rightarrow \frac{dy}{dx} &= \frac{3x^2 + 2x + 1}{3y^2} \\ &= \frac{3x^2+2x+1}{3 \sqrt[3]{(x+1)^2(x^2+1)^2}} \\\\ \Rightarrow \left. \frac{dy}{dx} \right\lvert_{x=0} &= \frac{1}{3}.\ _\square \end{align} \]